phân tích đa thức thành nhân tử: x4+2008x2 +2007x+2008
phân tích đa thức thành nhân tử:
x^4+2008x^2+2007x+2008
Phân tích đa thức thành nhân tử:
x^4 + 2008x^2 + 2007x + 2008
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x^4 + 2008x^2 + 2007x + 2008
\(=x^4+x^2+2007x^2+2007x+2007+1\)
\(=x^4+x^2+1+2007\left(x^2+x+1\right)\)
\(=\left(x^2+1\right)^2-x^2+2007\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+1\right)+2007\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2008\right)\)
Phân tích đa thức thành nhân tử:
2x^4+2008x^3+2007x+2008
Phân tích đa thức thành nhân tử : x4 + 2008x2 +2007x +2008
x^4+2008x^2+2007x+2008
=x^4+2008x^2+2008x-x+2008
=(x^4-x)+(2008x^2+2008x+2008)
=x(x^3-1)+2008(x^2+x+1)
=x(x-1)(x^2+x+1)+2008(x^2+x+1)
=(x^2+x+1)(x^2-x+2008)
Phân tích đa thức thành nhân tử
1/ x4+2008x2+2007x+2008
x4+2008x2+2007x+2008
<=> x4-x+2008x2+2008x+2008
<=> x(x3-1)+2008(x2+x+1)
<=> x(x-1)(x2+x+1)+2008(x2+x+1)
<=> (x2+x+1)(x2-x+2008)
Phân tich đa thức thành nhân tử
x4 + 2008x2 + 2007x + 2008
=x4+2008x2+2008x-x+2008
=(x4-x)+(2008x2+2008x+2008)
=x(x3-1)+2008(x2+x+1)
=x(x-1)(x2+x+1)+2008(x2+x+1)
=(x2+x++1)(x2-x+2008)
Phân tích đa thức thành nhân tử :
a, \(x^8+x^4+1\)
b, \(x^4+2008x^2+2007x+2008\)
a)\(x^8+2x^4+1-x^4=\left(x^4+1\right)^2-\left(x^2\right)^2\)
\(=\left(x^4+x^2+1\right)\left(x^4-x^2+1\right)\)
\(=\left(x^2-x+1\right)\left(x^2+x+1\right)\left(x^4-x^2+1\right)\)
\(=\left(x^4+x^3+x^2\right)-\left(x^3-2007x^2-2007x-2008\right)\)
\(=x^2\left(x^2+x+1\right)-\left[x\left(x^2+x+1\right)-2008\left(x^2-x-1\right)\right]\)
\(=x^2\left(x^2+x+1\right)-\left(x^2+x+1\right)\left(x-2008\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2008\right)\)
Câu \(\left(x^2+x+1\right)\left(x^2-x+2008\right)\) là câu b nha!Quên ghi đề
Phân tích đa thức thành nhân tử:
\(x^4+2008x^2+2007x+2008\)
Giải phương trình:
\(x^2\)-3x+2+ lx-1l
giải phương trình:
Nếu \(x\ge1\)phương trình trở thành : \(x^2-3x+2=x-1\Leftrightarrow x^2-4x+3=0\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}TM}\)Nếu \(x< 1\)\(\Rightarrow x^2-3x+2=1-x\Leftrightarrow x^2-2x+1=0\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1L\)VẬY NGHIỆM PHƯƠNG TRÌNH LÀ : x=1 hoặc x=3\(x^4+2008x^2+2007x+2008\)
\(=x\left[x\left(x^2+2008\right)+2007\right]+2008\)
\(=\left[\left(x-1\right)x+2008\right]\left(x^2+x+1\right)\)
\(=\left(x^2-x+2008\right)\left(x^2+x+1\right)\)
~(‾▿‾~)
\(x^4+2008x^2+2007x+2008\)
\(=x^4+2007x^2+x^2+2007x+2007+1\)
\(=\left(x^4+x^2+1\right)+\left(2007x^2+2007x+2007\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+1\right)+2007\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+1+2007\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2008\right)\)
Phân tích đa thức thành nhân tử:
a/\(x^2+7x+6\)
b/\(x^2+2008x^2+2007x+2008\)
E đg cần gấp
a.\(x^2+7x+6\)
\(=x^2+x+6x+6\)
\(=x\left(x+1\right)+6\left(x+1\right)\)
\(=\left(x+1\right)\left(x+6\right)\)
Sửa đề:.\(x^4+2008x^2+2007x+2008\)
\(=x^4+x^2+1+2007x^2+2007x+2007\)
\(=\left(x^4+x^2+1\right)+2007\left(x^2+x+1\right)\)
\(=\left(x^4+x^3+x^2-x^3-x^2-x+x^2+x+1\right)+2007\left(x^2+x+1\right)\)
\(=\left[x^2\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)\right]+2007\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+1\right)+2007\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2008\right)\)
Trả lời:
a, x2 + 7x + 6
= x2 + x + 6x + 6
= ( x2 + x ) + ( 6x + 6 )
= x ( x + 1 ) + 6 ( x + 1 )
= ( x + 6 ) ( x + 1 )